Question

Suppose the sum of the first $m$ terms of an arithmetic progression is $n$ and the sum of its first $n$ terms is $m$, where $m \ne n$. Then, the sum of the first $(m + n)$ terms of the arithmetic progression is

• A. $1 - mn$
• B. $mn - 5$
• C. $- (m + n)$
• D. $m + n$

Answer 1

Answer: (C)

Given, $S_m =n$ and $S_n = m$
$S_m = \frac{m}{2} [2a + ( m-1) d ] = n\,\,...(i)$
$S_n = \frac{n}{2}(2a+(n-1)d ) = m \,\,...(ii)$
On subtracting Eq. $(ii)$ from Eq. $(i)$, we get
$(m-n) a + ( m-n) (m+n - 1) \frac{d}{2}$
$= - (m-n)$
$\Rightarrow 2a + ( m +n -1) d = -2 \,\,[m\ne n]$
$\therefore s_{m+n} = \frac{m+n}{2} (2a + ( m+ n - 1)d)$
$ = \frac{m+n}{2} (-2) = - ( m + n)$